What is (81x^{-32}y^4)^{3/4} ?

The solution to (81x^{-32}y^4)^{3/4} is (27y^3)/(x^{24)}

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simplify (81x^{-32}y^4)^{3/4}

simplify

(81 x^{- 32} y^{4})^{\frac{3}{4}}

\frac{27 y ^{3}}{x ^{24}}

Solution steps

(81 x^{- 32} y^{4})^{\frac{3}{4}}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(81 x^{- 32} y^{4})^{\frac{3}{4}} = 8 1^{\frac{3}{4}} (x^{- 32})^{\frac{3}{4}} (y^{4})^{\frac{3}{4}}

= 8 1^{\frac{3}{4}} (x^{- 32})^{\frac{3}{4}} (y^{4})^{\frac{3}{4}}

8 1^{\frac{3}{4}} = 27

= 27 (x^{- 32})^{\frac{3}{4}} (y^{4})^{\frac{3}{4}}

Simplify

(x^{- 32})^{\frac{3}{4}} : \frac{1}{x ^{24}}

= 27 \cdot \frac{1}{x ^{24}} (y^{4})^{\frac{3}{4}}

Simplify

(y^{4})^{\frac{3}{4}} : y^{3}

= 27 \cdot \frac{1}{x ^{24}} y^{3}

Simplify

27 \cdot \frac{1}{x ^{24}} y^{3} : \frac{27 y ^{3}}{x ^{24}}

= \frac{27 y ^{3}}{x ^{24}}

(6 g)^{3}

5 w^{2} = 26 w + 24

e^{t} \cdot e^{t}

3 x^{2} - 12 x - 15

What is (81x^{-32}y^4)^{3/4} ?
The solution to (81x^{-32}y^4)^{3/4} is (27y^3)/(x^{24)}